SQL查询：将行转换为列

Question

这是我的桌子的一个例子。

我需要做一个查询，显示那些在两个月（11 或 12）中的一个或两个月中的一个费用为 0 的 ID。

因此，从示例中，我需要显示 ID 1、3、4 而不是 2，如下面的屏幕截图所示。

我尝试了以下查询：

SELECT 
    t1.id, t1.month, t1.fee, t2.id, t2.month, t2.fee
FROM
    table t1, table t2
WHERE t1.id = t2.id
  AND t1.month = '11'
  AND t2.month = '12'
  AND (t1.fee = 0 OR t2.fee = 0);

但是通过这个查询，我只看到 ID 1,3 而不是 ID 4。我猜这是因为 t1.id = t2.id 但不知道该怎么做。

Answer 1

您可以使用条件聚合。在 Postgres 中，这可以利用 filter 语法：

SELECT t.id,
       11 as month,
       MAX(t.fee) FILTER (WHERE t.month = 11) as fee_11, 
       12 as month,
       MAX(t.fee) FILTER (WHERE t.month = 12) as fee_12
FROM t
GROUP BY t.id
HAVING MAX(t.fee) FILTER (WHERE t.month = 11) = 0 OR
       MAX(t.fee) FILTER (WHERE t.month = 12) = 0;

注意：两个月份的列是多余的。

Answer 2

你需要条件聚合

  select id,month,max(case when month=11 then fee end) fee11,
     max(case when month=12 then fee end) as fee12
    from (
    select * from table t1
    where t1.id in ( select id from table where fee=0)
    ) a group by id,month

Answer 3

Sql ansi 兼容查询

    SELECT id, 
        MAX(CASE WHEN MONTH = 11 THEN MONTH ELSE NULL END) AS month11, 
        MAX(CASE WHEN MONTH = 11 THEN fee ELSE NULL END) AS fee11, 
        MAX(CASE WHEN MONTH = 12 THEN MONTH ELSE NULL END) AS month12, 
        MAX(CASE WHEN MONTH = 12 THEN fee ELSE NULL END ) AS fee12
    FROM t
    GROUP BY id
    HAVING ( MAX(CASE WHEN MONTH = 11 THEN fee ELSE NULL END) = 0 OR MAX(CASE WHEN MONTH = 12 THEN fee ELSE NULL END ) = 0 )
    ORDER BY id